Question

How many grams of AgNo3 are required to make 25 ml of .80M solution

Answer 1

3.39 grams of AgNo3 are required to make 25 ml of .80M solution

Step-by-step explanation:

Molarity , M = 0.80

Volume = 25 ml

`=(25)/(1000) L`

V = 0.025 L

`molarity = (moles)/(Volume\ of\ solution(L))`

`M=(n)/(V)`

Here n = number of moles

`n=M* V`

`n=0.8* 0.025`

n = 0.02 mole

`moles = (given\ mass)/(Molar\ mass)`

Molar mass of AgNO3 = 169.87 g/mol

`mass =(n)/(Molar\ mass)`

`mass=(0.02)/(169.87)`

Mass = 3.397 g